103 Trigonometry Problems

5. Solutions to Advanced Problems 195
where S k = a 1 + a 2 +···+a k , for k = 1, 2,...,n.
Proof:
as desired.
Set S 0 = 0. Then a k = S k − S k−1 for k = 1, 2,...,n, and so
n∑
a k b k =
k=1
n∑
(S k − S k−1 )b k =
k=1
∑n−1
= S n b n + S k b k −
k=1
n∑
S k b k −
k=1
n∑
S k−1 b k
k=1
n∑
S k−1 b k − S 0 b 1
k=2
∑n−1
∑n−1
= S n b n + S k b k − S k b k+1
k=1
k=1
∑n−1
= S n b n + S k (b k − b k+1 ),
k=1
Lemma 2 [Abel’s inequality] Let n be a positive integer, and let a 1 ,a 2 ,...,a n
and b 1 ,b 2 ,...,b n be two sequences of real numbers with b 1 ≥ b 2 ≥ ··· ≥
b n ≥ 0. Then
n∑
mb 1 ≤ a k b k ≤ Mb 1 ,
k=1
where S k = a 1 + a 2 +···+a k , for k = 1, 2,...,n, and M and m are the
maximum and minimum, respectively, of { S 1 ,S 2 ,...,S n }.
Proof:
1gives
Note that b n ≥ 0 and b k − b k+1 ≥ 0 for k = 1, 2,...,n− 1. Lemma
n∑
k=1
∑n−1
a k b k = S n b n + S k (b k − b k+1 )
k=1
n−1
∑
≤ Mb n + M (b k − b k+1 ) = Mb 1 .
k=1
establishing the second desired inequality. In exactly the same way, we can
prove the first desired inequality.
Lemma 3
Let x be a real number that is not an even multiple of π, then
n∑ sin kx
∣ k ∣ ≤ 1
(m + 1) ∣ ∣sin x ∣ ,
2
k=m+1